Question 712093: Solve by factorization
4^x-7.2^x+12=0
thanks
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! 
The first key to solving this equation is to recognize that one base with an exponent x, 4, is a power of the other base with an exponent, 2. So we can rewrite 4 as a power of 2:
The first term is a power of a power. The rule for the exponents is to multiply them:
The next key is to notice that the exponent of the first 2 is exactly twice the exponent of the other 2. An equation like this is called an equation in "quadratic form". Quadratic form equations can be solved in much the same way as regular quadratic equations.
If you have trouble seeing the "quadratic-ness" of the equation it can be helpful to use a temporary variable. Set the temporary variable equal to the base and the smaller exponent. So:
Let 
Then 
Substituting these into the equation we get:
This is obviously a quadratic equation. We can solve it by factoring:
From the Zero Product Property:
 or 
Solving these we get:
 or 
Of course we are not interested in solutions for q. We want solutions for x. So we substitute back in for q:
 or 
To solve for x we have some more work to do. I hope the solution to the first equation is obvious, x=2. If not, then you can solve it in a way similar to the way we will solve the second equation.
Since we do not know what power of 2 results in a 3, we will need to use logarithms to solve for x. The logarithm we use can be of any base. But choosing certain bases have advantages:- Choosing the logarithm's base to match the base of the exponent (where the variable is) will result in the simplest possible expression for the exact solution.
- Choosing the logarithm's base to match a base our calculators "know" (base 10, "log", or base e, "ln") will result in an exact expression which can easily converted to a decimal approximation is wanted/needed.
I'll do it both ways so you can see the difference.
Matching the logarithm's base to the exponent's base, we'll use base 2 logs:
Next we use a property of logarithms,  , which allows us to move the exponent of the argument of a log out in front as a coefficient. (It is this very property of logarithms which is the reason we use logarithms. It allows us to, in effect, change an exponent (where the variable is) into a coefficient (where we can then "get at" the variable with "regular" algebra.) Using this property we get:
By definition,  . (This is why matching the base of the log to the base of the exponent results in the simplest expression.) So the left side becomes:
This is an exact expression for the second solution to your equation. (The first solution was x = 2.)
P.S. Using a base that our calculators "know". I'll use base e, "ln". The steps are mostly the same so I'll omit commentary except to explain differences:
This time the log on the left is not a 1. Dividing by the log on the left:
Although it looks quite different from the solution we got using base 2 logs,  , this is also an exact expression for the second solution to your equation. By using ln's we get an expression which is not as simple as the other but which can be easily entered into our calculators to get a decimal approximation if wanted/needed.
P.P.S. Once you've gotten some experience with these quadratic form equations you will no longer need to use a temporary variable. You will be able to see how to do directly from:
to
to
 or 
etc.
|
|
|
